Question 4·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

A circle inscribed in the square touches each side at its midpoint. The distance from the square’s center to a side is half of $40$, so the cone’s base radius is $r=\frac{40}{2}=20$ inches. This is also the center-to-midpoint distance needed for the height triangle.

The apex of a right pyramid is above the base center. Its perpendicular height $h$ meets the $20$-inch center-to-midpoint segment at a right angle. The $29$-inch slant height is the hypotenuse, so the Pythagorean theorem gives $h^2+20^2=29^2$. Volume uses the perpendicular height, not the slant height.

Type $h_{1}^{2}+20^{2}\sim29^{2}\left\{h_{1}>0\right\}$ in Desmos. Use $h_1$ instead of $h$ and $\sim$ instead of $=$ so Desmos solves for the unknown; the restriction keeps the length positive. Under PARAMETERS, Desmos shows $h_1=21$, so the perpendicular height is $21$ inches.

For a pyramid, $V=\frac13Bh$, where $B$ is base area and $h$ is perpendicular height. Its square base has area $40^2$, so type $\frac13(40)^2(21)$. Desmos prints $11200$, the whole pyramid’s volume in cubic inches.

For a cone, $V=\frac13\pi r^2h$. The cone has the same apex and base plane as the pyramid, so its perpendicular height is also $21$. Type $\frac13(20)^2(21)$ for the coefficient of $\pi$. Desmos prints $2800$, making the cut-out volume $2800\pi$ cubic inches.

The cavity removes the cone’s volume from the pyramid’s volume, so $V_{\text{remaining}}=11200-2800\pi$. The material remaining has $11200-2800\pi$ cubic inches of volume. Choice C.